Prove that then tan-16x-8x31-12x2-tan-14x1-4x2-tan-12x-2x- 1-3-

Consider the left hand sideL.H.S =tan^ 1(6x 8x^31 12x^2) tan^ 1(4x1 4x^2) We know that tan^ 1A tan^ 1B=tan^ 1(A B1+AB) L.H.S =tan^ 1(6x 8x^31 12x^2 ...

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Byju's Answer

Standard XII

Mathematics

Sign of Trigonometric Ratios in Different Quadrants

Prove that th...

Question

Prove that then ${tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}$ .

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Solution

Consider the left hand side
L.H.S $= {tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}})$
We know that ${tan}^{- 1} A - {tan}^{- 1} B = {tan}^{- 1} (\frac{A - B}{1 + A B})$
L.H.S $= {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{6 x - 8 x^{3}}{1 - 12 x^{2}} - \frac{4 x}{1 - 4 x^{2}}}{1 + (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) (\frac{4 x}{1 - 4 x^{2}})} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠$
$= {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{\frac{(6 x - 8 x^{3}) (1 - 4 x^{2}) - 4 x (1 - 12 x^{2})}{(1 - 12 x^{2}) (1 - 4 x^{2})}}{\frac{(1 - 12 x^{2}) (1 - 4 x^{2}) + 4 x (6 x - 8 x^{3})}{(1 - 12 x^{2}) (1 - 4 x^{2})}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠$
$= {tan}^{- 1} (\frac{(6 x - 8 x^{3}) (1 - 4 x^{2}) - 4 x (1 - 12 x^{2})}{(1 - 12 x^{2}) (1 - 4 x^{2}) + 4 x (6 x - 8 x^{3})})$
$= {tan}^{- 1} (\frac{6 x - 24 x^{3} - 8 x^{3} + 32 x^{5} - 4 x + 48 x^{3}}{1 - 4 x - 12 x^{2} + 48 x^{4} + 24 x^{2} - 32 x^{4}})$
$= {tan}^{- 1} (\frac{32 x^{5} + 16 x^{3} + 2 x}{16 x^{4} + 8 x^{2} + 1})$
$= {tan}^{- 1} (\frac{2 x (16 x^{4} + 8 x^{2} + 1)}{16 x^{4} + 8 x^{2} + 1})$
$= {tan}^{- 1} 2 x$
Thus, L.H.S $=$ R.H.S

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Similar questions

Q. Prove that

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

Q. Prove that :

t a n^{- 1} [\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}] - t a n^{- 1} [\frac{4 x}{1 - 4 x^{2}}] = t a n^{- 1} 2 x, | 2 x | < \frac{1}{\sqrt{3}}

Q. i) Solve for

x : {tan}^{- 1} (x - 1) + {tan}^{- 1} x + {tan}^{- 1} (x + 1) = {tan}^{- 1} 3 x

ii) Prove that

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}

Q. The expression

{tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}})

in the interval

| 2 x | < \frac{1}{\sqrt{3}}

simlifies to:

{tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + {tan}^{- 1} (\frac{2 + 3 x}{3 - 2 x})

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Prove that then tan−1(6x−8x31−12x2)−tan−1(4x1−4x2)=tan−12x;|2x|<1√3.

Prove that then ${tan}^{- 1} (\frac{6 x - 8 x^{3}}{1 - 12 x^{2}}) - {tan}^{- 1} (\frac{4 x}{1 - 4 x^{2}}) = {tan}^{- 1} 2 x; | 2 x | < \frac{1}{\sqrt{3}}$ .